Ijeas0205053

International Journal of Engineering and Applied Sciences (IJEAS) ISSN: 2394-3661, Volume-2, Issue-5, May 2015

Optimize Renting Times of Machines in Flow-Shop Scheduling Laxmi Narain 

Abstract— This paper studies three-machine scheduling problems in the situation when one has got the assignment but does not have one's own machines and has to take machines on rent to complete the assignment. Minimization of total rental cost of machines may be the criterion in this type of situation.

Here, we have considered a rental policy in which second and third machines will not be taken on rent at times when the first job is completed on first and second machines respectively but these machines will be taken on rent subject to some criterion.

The objective is: for a given sequence obtain the latest times at which the machines should be taken on rent so that total rental cost is minimum without altering the total elapsed time. We have obtained a simple and efficient algorithm, without using Branch-and-Bound technique. Numerical example is given to illustrate the algorithm.

Index Terms— Flow-shop, Scheduling, Idle Time, Completion Time, Elapsed Time, Rental Time, Rental Cost.

I. INTRODUCTION In flow-shop problem, situation can occur in practice when one has got the assignment but does not have one's own machines or does not have enough money for the purchase of machines, under these circumstances, may take machines on rent to complete the assignment. Minimization of total rental cost of machines will be the criterion in these types of situations.

The following renting policies generally exist: Policy I: All the machines are taken on rent at one time and are returned also at one time.

Policy II: All the machines are taken on rent at one time and are returned as and when they are no longer required.

Sen [6] surveyed the bicriteria scheduling research for a singe machine. Chandersekhran [5] gave a technique based on Branch-and-Bound method and satisfaction of certain conditions to obtain a sequence which minimizes total flow-time subject to certain conditions which are to be satisfied. Bagga and Ambika [4] provided the procedure for obtaining sequence(s) in n-job, m-machine special flow-shop problems which gives minimum possible makespan while minimizing total flow-time. Narain and Bagga [11] studied n-job, m-machine special flow-shop problems which give minimum possible mean flowtime while minimizing total elapsed time. Narain and Bagga [8] determine the sequence which minimizes the total elapsed time subject to zero total idle time of machines i.e., machines should not remain idle once they start the first job. Narain and Bagga [10] studies n-job, m-machines flowshop problems when processing times of jobs on various machines follow certain conditions and the objective is to obtain a sequence which minimizes total elapsed time under no-idle constant. Narain and Bagga [9] studied n-job, 2-machine flowshop problem and provided an algorithm for obtaining a sequence which gives minimum possible mean flowtime under no-idle constraint.

This paper studies bi-criteria in three-machine flow-shop problems under rental Policy III. In this paper, Policy III is modified. Here second and third machines will not be taken on rent at times when the first job is completed on first and second machines respectively but these machines will be taken on rent subject to minimum total elapsed. The objective is: Obtain the time at which machines should be taken on rent so that total rental cost as minimum as possible without altering the total elapsed time. For any sequence S, Total rental cost of machines 3

=

n

[ p

i, j

( S )  Ii , j ( S )]  cj

j 1 i 1

Policy III: All the machines are taken on rent as and when they are required and are returned as and when they are no longer required for processing.

Bagga [1] studied three-machine problem under policy P1 and provide the sequence to minimize the total rental cost of machines. Under P2; for three-machine flow-shop problem, Bagga and Ambika [2] provided a Branch-and-Bound algorithm.

In this paper we are considering rental policy in bi-criteria scheduling problems A survey of scheduling literature has revealed the desirability of an optimal schedule being evaluated by more than one performance measures or criteria.

Various authors [3-16] have studied the flow-shop problems having more than one optimization measures. Gupta and Dudek [7] strongly recommended the use of combination of criteria total flow-time and total elapsed time. Dileepan and LAXMI NARAIN, Associate Professor, Department of Mathematics, Acharya Narendra Dev College, University of Delhi, Delhi, India.

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Where pi,j(S) is the processing time of ith job of sequence S on machine Mj, Ii,j(S) is the idle time of machine Mj for ith job of sequence S and Cj is rental cost per unit time of machine Mj. Here, the processing times pi,j(S) and rental cost Cj(S) are constant. Therefore, we can only reduce idle times Ii,j(S). To reduce idle times on machines, we delay the times of renting of machines to process jobs. We have obtained a simple and efficient algorithm to provide the times at which machines should be taken on rent so that total rental cost as minimum as possible without altering the total elapsed time.

Numerical example is given to illustrate the algorithm.

II. MATHEMATICAL FORMULATION Notations: S: Mj: pi,j(S):

Sequence of jobs 1, 2, …, n.

Machine j; j=1, 2, 3.

Processing time of ith job of sequence S on machine Mj.

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OPTIMIZE RENTING TIMES OF MACHINES IN FLOW-SHOP SCHEDULING

Idle time of machine Mj for ith job of sequence S. Rental cost per unit time of machine Mj. The time when Mj is taken on rent for sequence S. Completion time of ith job of sequence S on machine Mj. Z′i,j(S): Completion time of ith job of sequence S on machine Mj when Mj starts processing jobs at time Hj(S).

T2(S): Total time for which M2 is required when M2 starts processing jobs at time H2(S). i = 1, 2, …, n and j= 1, 2, 3.

Ii,j(S): Cj: Hj(S): Zi,j(S):

n

Lemma 2.1: If M3 starts processing jobs at time H3 = then H3 ≥ Z1,2 and Z′k,3 ≥ Zk,2 for k > 1. n

Proof: H3

n

= Z1,2 +

I

i ,3

≥ 0, therefore, H3 ≥ Z1,2

i ,3

i 2

Now, Ik,3 = max(Zk,2 - Zk-1,3 , 0) Therefore, Ik,3 ≥ Zk,2 - Zk-1,3 i.e., Zk-1,3 + Ik,3 ≥ Zk,2 k 1

i.e.,

p

i ,3

I

p

i ,3

i 1 k

i.e.,

k 1

 Ii,3 + i ,3 +

i 1

1

+ Ik,3 ≥ Zk,2

i 1 k 1

≥ Zk,2

i 1

n

p1,3

Since,

i 1

I

i ,3 k

Therefore,

p

 Ii,3 + i 1

m 1

 Ii,3 + i 1 m

= max (Zm+1,2,

I

i ,3

+

i 1

) + pm+1,3

i.e., H3 +

+

i  k 1

p

i ,3

≥ Zk,2

i 1

k 1

p

i ,3

≥ Zk,2

i 1

p

i ,3

≥ Zk,2

m

i.e., Z′k-1,3 ≥ Zk,

i 1 m

Hence, this lemma is proved.

 pi,3 ) + pm+1,3 p

i ,3

+ Im+1,3) + pm+1,3

Theorem 2.3: Total elapsed time will not be altered, if M2 starts processing jobs at time H2 = min {Yk}, where Y1 = H3 - p1,2 and

i 1

k

Yk = Z′k-1,3 -

p

i,2

; k=2, 3, …, n

i 1

Proof: H2 = Yr= min {Yk}; k=1, 2, …, n For k = 1; Yr= min {Yk}; k=1, 2, …, n Therefore, Yr ≤ Y1 i.e., Yr + p1,2 ≤ Y1 + p1,2 i.e., Yr + p1,2 ≤ H3 … (1) From Lemma 3.1; Z1,2 ≤ H3 … (2) Now, Z′1,2 = max (Yr + p1,2 , Z1,2) From equations (1) and (2); Z′1,2 ≤ H3 … (3) For k > 1; Yr= min {Yk}; k=2, 3, …, n

Hence, by mathematical induction this theorem holds for all k, where k = 1, 2, ..., n.

If M3 starts processing jobs at time H3 , where H3 = Zn,3 n

, then total elapsed time Zn,3 is not altered and M3 is

i 1

engaged for minimum time equal to sum of the processing times of all the jobs on M3. Moreover, it can be easily shown that if M3 starts processing jobs at time H3, then k

p

i ,3

i 1

= max (Zm+1,2, Zm,3 + max (Zm+1,2- Zm,3 , 0)) + pm+1,3 = max (Zm+1,2, max (Zm+1,2, Zm,3 )) + pm+1,3 = max (Zm+1,2, Zm,3 ) + pm+1,3 = Zm+1,3 Therefore, the result holds for k = m+1 also.

Z′k,3 = H3 +

I

k 1

i ,3

i 1

= max (Zm+1,2,

i ,3 +

n

i.e.,

k 1

n

I i 1

m

= max (Zm+1,2, H3 +

≥ 0,

i  k 1

= p1,1 + p1,2 + p1,3 = Z1,3 Therefore, the result holds for k = 1.

Let the result holds for k = m For k = m+1; Z′m+1,3 = max (Zm+1,2, Z′m,3 ) + pm+1,3

i ,3

I

n

Since,

Proof: Let Z′i,3 be the completion time of ith job on machine M3 when M3 starts processing jobs at time H3. The proof of the theorem is based on the method of mathematical induction.

For k = 1; Z′1,3 = H3 + p1,3

p

i ,3

i 2

then Zk,3 will remain unaltered.

i ,3 +

I i 2 n

i 1

I

i ,3

= I1,3 +

k

=

I

=

i 1

Theorem 2.1: If we start processing jobs on M3 at time H3 = i ,3 ,

i ,3 ,

i 1

Let n jobs require processing over three machines M1, M2 and M3 in the order M1→ M2→ M3.

I

I

i ,3

i 1

85

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International Journal of Engineering and Applied Sciences (IJEAS) ISSN: 2394-3661, Volume-2, Issue-5, May 2015 Therefore, Yr ≤ Yk ; k=2, 3, …, n k

i.e., Yr +

p

p

≤ Yk +

i,2

i 1 k

i.e., Yr +

p

M3 will start processing job r at time = max (Z′r,2, Z′r-1,3) = Z′r,2 … (10)

k

n

i,2

Therefore, total elapsed time ≥ Z′r,2 +

i 1

p

i ,3

i 1

≤ Z′k-1,3

i,2

… (4)

n

≥ Z′r-1,3 +

i 1

p

i ,3

= Z′n,3 = Zn,3

i 1

From Lemma 2.1; Zk,2 ≤ Z′k-1,3 Now,

… (5)

Hence, total elapsed time will increase if M2 starts processing jobs at time H2 > min {Yk}; k=1, 2, …, n.

k

Z′k,2 = max (Yr +

p

i,2 ,

Zk,2)

By Theorem 2.1; the starting of processing jobs at time H3 on M3 will reduce the idle time of M3 to zero and M3 will be required only for time equivalent to the sum of the processing times of all the jobs on it. Therefore, total rental cost of M3 will be minimum (least). Total rental cost of M1 will always be minimum (least), since idle time of M1 is always zero.

Therefore, the objective is to minimize the rental cost of machine M2.

i 1

From equation (4) and (5); Z′k,2 ≤ Z′k-1,3; k=2, 3, …, n … (6) Taking k = n in equation (6); Z′n,2 ≤ Z′n-1,3 … (7) Total elapsed time = max (Z′n,2 , Z′n-1,3) + pn,3 = Z′n-1,3 + pn,3 = Z′n,3 = Zn,3 Hence, total elapsed time will not be altered if M2 starts processing jobs at time H2 = min {Yk};k=1, 2, …, n.

Theorem 2.4: Total elapsed time will increase, if M2 starts processing jobs at time H2 > min {Yk}, where Y1 = H3 - p1,2 and k

Yk = Z′k-1,3 -

p

i,2 ;

n

H3 = Zn,3(S) -

p i 1

i ,3

Yk(S) = H3 +

= Z′n,3 = Zn,3

i 1

Case 2: Let Yr= min {Yk}; k=1, 2, …, n Let H2 = Yk, then > Yr +

i 1

p

i,2

… (8)

p

i ,1 × C1+(Zn,2(S)- H2)×C2

+ (Zn,3(S) - H3)×C3

i 1

IV. EXAMPLE

r

i,2 ,

Zr,2)

Example 4.1: Consider the 10-Job, 3-Machine flow-shop problem with processing times in hours as given in Table 1.

The rental costs per unit time for machines M1, M2 and M3 are Rs. 50, Rs. 100 and Rs. 75 per hour respectively. Jobs are processed in the sequence 1-2-3-4-5-6-7-8-9-10.

i 1

r

Therefore, Z′r,2 ≥ Yk +

i 1

n

R(S) =

i 1

p

i 1

Step 5: Compute rental time H2 of M2 for sequence S H2 = min {Yk}; k=1, 2, …, n.

Now, Z′r,2 = max (Yk +

k

 pi,3(S ) -  pi,2(S ) ;

Step 6: Compute total rental cost for sequence S

r

i,2

k 1

k=2, 3, …, n

Hence, total elapsed time will increase if M2 starts processing jobs at time H2 > Y1.

p

(S )

i ,3

Step 4: For sequence S, compute Y1(S) = H3 - p1,2(S)

i ,3

n

p

p i 1

n

Therefore, total elapsed time ≥ H2 + p1,2 +

Yk +

Algorithm 3.1: Step 1: Let S be the given sequence.

Step 3: Compute rental time H3 of M3 for sequence S

Proof: There arise two cases: Case 1: H2 > Y1, then H2 + p1,2 > Y1 + p1,2 = H3 i.e., H2 + p1,2 > H3

r

III. ALGORITHM

Step 2: Compute Zn,2(S) and Zn,3(S).

k=2, 3, …, n

i 1

≥ H3 +

The following algorithm provides the procedure to determine the times at which machines should be taken on rent to minimize the total rental cost without altering the total elapsed time.

p

i,2

i 1

From equation (8); r

Z′r,2 > Yr +

p

i,2

Table 1: Processing Times of Jobs on Machines

= Z′r-1,3

i 1

i.e., Z′r,2 > Z′r-1,3

… (9)

Jobs

86

M1

Machines M2

M3

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OPTIMIZE RENTING TIMES OF MACHINES IN FLOW-SHOP SCHEDULING

1 2 3 4 5 6

2 4 9 5 5 15

2 5 3 7 11 5

3 7 4 12 5 6

7

10

2

5

8

4

5

2

9

6

3

4

10

7

1

1

3

Y4 = H 3 +

i 1

Machines M2

In-Out

In-Out

In-Out

0-2 2-6 6-15 15-20 20-25 25-40

2-4 6-11 15-18 20-27 27-38 40-45

4-7 11-18 18-22 27-39 39-44 45-51

7

40-50

50-52

52-57

8

50-54

54-59

59-61

9

54-60

60-63

63-67

10

60-67

67-68

68-69

REFERENCES [1] Bagga, P.C., ―Sequencing in a rental situation‖, Jr. of Canadian Operations Research Society, vol. 7, 1969, pp. 152-153.

[2] Bagga, P.C. and Ambika Bhambani, ― Minimizing rental costs in three-machine sequencing problem‖, Jr. of Indian Association for Productivity, Quality and Reliability, vol. 21, 1996, pp.

73-77.

[3] Bagga, P.C. and Ambika Bhambani, ― Bicriteria in flowshop scheduling problems‖, Journal of Combinatorics, Information and System Sciences, vol. 21, no. 3-4, 1996.

[4] Bagga, P.C. and Ambika Bhambani, ―Bicriteria in special flowshop problems‖, Indian Journal Pure and Applied Mathematics, vol.

33, no. 4, 2002, pp. 435- 441.

[5] Chandersekharan Rajendran, ―Two-stage production scheduling problem with bicriteria‖, Operational Research Society, U. K, vol. 43, no. 9, 1992, pp. 871-884.

[6] Dileepan, P. and Sen, T, ―Bicriterion state scheduling research for a single machine‖, Omega, vol. 16, 1988, pp. 53-59.

[7] Gupta, J.N.D. and Dudek, R.A, ―Optimality criteria for flowshop schedule‖, AIIE Transactions, vol. 3, no. 3, 1971, pp. 199-205.

[8] Narain, L. and Bagga, P.C, ―Minimizing total elapsed time subject to zero idle time of machines in n3 flow-shop problem‖, Indian Journal of Pure & Applied Mathematics, vol. 34, 2003, pp. 219-228.

[9] Narain, L. and Bagga, P. C., ―Flow-shop/no-idle scheduling to minimize total elapsed time‖, Journal of Global Optimization, vol. 33, 2004, pp. 349-367.

i ,3

= 69 – 49 = 20 Therefore, machine M3 should be taken on rent after 20 hours of starting the processing of the first job on machine M1 (Step 3).

For sequence S, Y1 = H3 - p1,2 = 20 -2 = 18 1

 pi , 3 i 1

2

p

i,2

i 1

= 20 + 3 – 7 = 16 2

Y3 = H 3 +

 pi , 3 i 1

3

p

C1 + (Zn,2(S) - H2) × C2 + (Zn,3(S) - H3) × C3

In this paper, we have proved three theorems to find out the times at which machines should be taken on rent so that total elapsed do not change when we delay the processing of jobs on machines. A simple and efficient algorithm is developed by using these theorems which provide the times at which machines should be taken on rent so that total rental cost is as minimum as possible without altering the total elapsed time.

i 1

Y 2 = H3 +

i ,1 ×

V. CONCLUSION

10

p

p

= 67 × 50 + (68-16) × 100 + (69-20) × 75 = 3350 + 52 × 100 + 49 × 75 = 3350 + 5200 + 3675 = 12225 (Step 6) Hence, for sequence S = 1-2-3-4-5-6-7-8-9-10 the minimum total rental cost is Rs. 12,225 without altering the total elapsed time (69 hours) when machine M1 is taken on rent in the starting of processing the jobs, M2 after 16 hours of starting the processing of first job on machine M1 and M3 after 20 hours of starting the processing of first job on machine

Thus, the completion time Z10,2 = 68 hours and Z10,3 = 69 hours (Step 2).

The rental time H3 of machine M3 for sequence S is H3 = Z10,3 -

i,2

i 1

i 1

M3

1 2 3 4 5 6

p

n

R(S) =

Table 2: Completion Times In-Out for Sequence S

M1

4

= 20 + (3+7+4) – (2+5+3+7) = 20 +14 – 17 = 17 Continuing in this way, Y5 = 18; Y6 = 18; Y7 = 22; Y8 = 22; Y9 = 21 and Y10 = 24 (Step 4).

Rental time H2 of machine M2 for sequence S is H2 = min {Yk} = min {18, 16, 20, 17, 18, 18, 22, 22, 21, 24} = 16 Therefore, machine M2 should be taken on rent after 16 hours of starting the processing of the first job on machine M1 (Step 5).

Total rental cost of machines for sequence S is

Applying Algorithm 3.1; The given sequence S = 1-2-3-4-5-6-7-8-9-10 (Step 1). For determining the completion time of last job on machines M2 and M3, this sequence is enumerated and its completion time In-Out is given in Table 2.

Jobs

 pi , 3 -

i,2

i 1

= 20 + (3+7) – (2+5+3) = 20 +10 – 10 = 20

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International Journal of Engineering and Applied Sciences (IJEAS) ISSN: 2394-3661, Volume-2, Issue-5, May 2015 [10] Narain, L. and Bagga, P. C., ―Flow-shop/no-idle scheduling to minimize mean flow-time‖, ANZIAM. Journal, vol. 47, 2005, pp.

265- 275.

[11] Narain, L. and Bagga, P. C., ―Bi-criteria in n × m flow-shop problems‖ IAPQR Transactions, vol. 31. 2006, pp. 57-71.

[12] Sen, T. and Gupta, S.K., ―A branch and bound procedure to solve a bi-criterion scheduling problem‖, AIIE Transactions, vol. 15, 1983, pp.

84- 88.

[13] Sen, T. and Dileepan, P., ―A bicriterion scheduling problem involving total flowtime and total tardiness‖, Journal of Information and Optimization Sciences, vol. 20, no. 2, 1999, pp. 155-170 [14] Sen, T. and Raiszadeh, F., ―An algorithm to minimize total flowtime and maximum job lateness in the two-machine flowshop system‖, Journal of Information and Optimization Sciences, vol. 18, no. 1, 1997, pp. 9-16.

[15] Sen, T., Raiszadeh, F. and Dileepan, P., ―A branch-and-bound approach to the bicriterion scheduling problem involving total flowtime and range of lateness‖, Management Science, vol. 34, no. 2, 1988, pp. 254-260.

[16] Van Wassenhove, L. N. and Gelders, L. F., ―Solving a bicriterion scheduling problem‖, European Journal of Operational Research, vol. 4, 1980, pp. 42- 48.

[17] Van Wassenhove, L. N. and Baker, K. R., ―A bicriterion approach to time/cost trade-off in sequence‖, European Journal of Operational Research, vol. 11, 1982, pp. 48-54.

AUTHOR’S BIOGRAPHY Dr laxmi Narain is currently an Associate Professor in Department of Mathematics, Acharya Narendra Dev College, University of Delhi, Delhi, India. He has specialization in Optimization Technique. He has done Ph.D. in Flow-shop Sequencing Problem from Department of Mathematics, University of Delhi under the Guidance of Dr. P. C. Bagga. He has published research papers in National and International Journals.

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